Let α=(1101)\alpha = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}α=(1011) . Then α−1Rα\alpha^{-1}R\alphaα−1Rα consists of the matrices
We see easily that the set of these matrices is exactly AAA . Therefore, AAA is just a "conjugate" of the subring RRR in the ring M2(k)\mathbf{M}2(k)M2(k) . In particular, A∼RA \sim RA∼R .